Question 3.22: The industrial process for making the ammonia used in fertil...
The industrial process for making the ammonia used in fertilizer, explosives, and many other products is based on the reaction between nitrogen and hydrogen at high temperature and pressure:
N_{2}(g) + 3 H_{2}(g) → 2 NH_{3}(g)
If 18.20 kg of NH_{3} is produced by a reaction mixture that initially contains 6.00 kg of H_{2} and an excess of N_{2}, what is the percent yield of the reaction?
Collect and Organize We know that the actual yield of NH_{3} is 18.20 kg. We also know that the reaction mixture initially contained 6.00 kg of H_{2} and that H_{2} must be the limiting reactant because the problem specifies the presence of excess N_{2}. We need to find the percent yield.
Analyze We need to use the mass of H_{2} to calculate how much NH_{3} could have been produced— the theoretical yield. We can then use the theoretical yield and the actual yield to calculate percent yield. We need the molar masses of H_{2} and NH_{3} and the stoichiometry of the reaction, which tells us that 2 moles of NH_{3} are produced for every 3 moles of H_{2} consumed.
Solve The molar masses we need are:
\mathscr{M}_{H_{2}} = 2(1.008 g/mol) =2.016 g/mol
\mathscr{M}_{NH_{3}} =14.01 g/mol + 3(1.008 g/mol) = 17.03 g/mol
We calculate the theoretical yield of NH_{3}:
6.00 \sout{kg H_{2}}\times \frac{10^{3} \sout{g H_{2}}}{1 \sout{kg H_{2}}} \times \frac{1 \sout{mol H_{2}}}{2.016 \sout{g H_{2}}} \times \frac{2 \sout{mol NH_{3}}}{3 \sout{mol H_{2}}} \times \frac{17.03 \sout{g NH_{3}}}{1 \sout{mol NH_{3}}} \times \frac{1 kg NH_{3}}{10^{3} \sout{g NH_{3}}} =33.8 kg NH_{3}
Then we divide the actual yield by this theoretical yield to determine the percent yield:
\frac{18.20 \sout{kg NH_{3}}}{33.8 \sout{kg NH_{3}}} \times 100 \%=53.8\%
Think About It A yield of about 54% may seem low, but it may be the best that can be achieved for a particular process. A great deal of chemical research goes into trying to improve the percent yield of industrial chemical reactions.